How do you integrate #int e^(2x)/sqrt(-e^(2x) +25)dx# using trigonometric substitution?

Question

1651 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-11T12:48:10

#=-5cos(sin^-1(e^x/5))+C#

#inte^(2x)/sqrt(-e^(2x)+25)dx#

#5sintheta=e^x#
#5costheta*d(theta)=e^xdx#

#inte^(2x)/sqrt(-e^(2x)+25)dx#=#int(5sintheta*5costheta)/sqrt(25-25sin^2theta)d(theta)#

#25-25sin^2theta=25cos^2theta#

simplify,

#int(5sintheta*5costheta)/sqrt(25-25sin^2theta)d(theta)#=#5intsinthetad(theta)#

#=-5costheta+C#

reverse the trigonometric substitution

#theta=sin^-1(e^x/5)#

so your answer will be:

#=-5cos(sin^-1(e^x/5))+C#